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Thermal Properties of Matter

JEE Main level, numerical answer, medium

95 of 126

Question 95 of 126

As per the given figure, two plates $A$ and $B$ of thermal conductivity $K$ and $2K$ are joined together to form a compound plate. The thickness of plates are $4.0cm$ and $2.5cm$ respectively and the area of cross-section is $120{cm}^{2}$ for each plate. The equivalent thermal conductivity of the compound plate is $\left(1+\dfrac{5}{\alpha }\right)K$ , then the value of $\alpha$ will be _____ .

Show answer and solution

Answer: 21

The thermal resistance of rod 1 will be

{R}_{1}=\dfrac{{l}_{1}}{KA}=\dfrac{4}{KA}

and rod 2 will be

{R}_{2}=\dfrac{{l}_{2}}{2KA}=\dfrac{2.5}{2KA}

.
Since they are connected in series,

{R}_{eq}={R}_{1}+{R}_{2}=\dfrac{10.5}{2KA}

.
If equivalent thermal conductivity is

{K}_{eq}

, then

{R}_{eq}=\dfrac{{l}_{1}+{l}_{2}}{{K}_{eq}A}=\dfrac{6.5}{{K}_{eq}A}

. Therefore,

\dfrac{10.5}{2KA}=\dfrac{6.5}{{K}_{eq}A}\Rightarrow {K}_{eq}=\dfrac{26}{21}K=\left(1+\dfrac{5}{21}\right)K

Hence,

\alpha =21

JEE Main question paper PDF

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JEE Main 2025 Apr 2 Shift 1 question paper

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